Question

If ${\int }_0^{\pi }xf\left(\sin x\right)dx=A{\int }_0^{\pi /2}f\left(\sin x\right)dx$ then A equals

• A. $\frac{\pi }{4}$
• B. $\pi$
• C. $0$
• D. $2\pi$

Answer 1

Answer: (B)

$\pi$

${\int }_0^{\pi }xf\left(\sin x\right)dx=A{\int }_0^{\frac{\pi }{2}}f\left(\sin x\right)dx$
or $A{\int }_0^{\frac{\pi }{2}}f\left(\sin x\right)dx$
$=\frac{\pi }{2}{\int }_0^{\pi }f\left(\sin x\right)dx={\int }_0^{\pi }xf\left(\sin x\right)dx$
$\Rightarrow$ $A{\int }_0^{\frac{\pi }{2}}f\left(\sin x\right)dx=\frac{\pi }{2}\times 2{\int }_0^{\frac{\pi }{2}}f\left(\sin x\right)dx$
$\Rightarrow$ $A{\int }_0^{\frac{\pi }{2}}f\left(\sin x\right)dx=\pi {\int }_0^{\frac{\pi }{2}}f\left(\sin x\right)dx$
$\Rightarrow$ $A=\pi$
Ans: B